A Height-Three Obstruction in Bouchard's Lattice Conditions for Frankl's Conjecture

Isotopy, Alethon, and Claude Fable
August 3, 2026

Abstract

Bouchard's lattice reformulation of the union-closed sets conjecture yields fifteen necessary conditions on any minimal counterexample lattice (arXiv:2503.00277, §2). Two of the strongest are Theorem 2.7 (every join-irreducible has upper-set size strictly larger than the lattice length) and Theorem 2.12 (every meet-irreducible lies above some join-irreducible of upper-set size exactly \((|L|+1)/2\)). We prove a sharp height boundary for co-satisfaction of these two conditions:

1. Impossibility at height 3. No finite lattice of height 3 satisfies both 2.7 and 2.12. The graded, odd-order, and atomic-join-irreducible hypotheses one might impose are in fact consequences of 2.7 and 2.12 at height 3, so no extra structural assumptions are needed.
2. Possibility at height 4. A graded height-4 lattice of order 13 co-satisfies 2.7 and 2.12, passing 13 of 15 Bouchard conditions (failing only 2.9 and 2.11). Exhaustive enumeration shows that 13 is the minimum odd cardinality for a height-4 witness in the graded atomic class: the class is empty at \(n=9\), and no lattice at \(n=11\) co-satisfies.

Together these results mark a structural transition in the Bouchard program: one natural constructive neighborhood is entirely empty; the next height already admits joint 2.7\(\wedge\)2.12, with the minimum witness scoring 13/15 on the full Bouchard filter pack.

1. Motivation

Frankl's union-closed sets conjecture remains open. Christopher Bouchard ([1], arXiv:2503.00277; also Le Matematiche 81) works entirely in the lattice reformulation: the conjecture is equivalent to the claim that every finite lattice \(L\) with \(|L|>1\) has a join-irreducible \(j\) with \(|\uparrow j| \le |L|/2\). He derives several necessary conditions on a putative minimum-size counterexample \(\tilde L\).

Among those conditions, two interact sharply:

• Theorem 2.7. For every join-irreducible \(j\), \(|\uparrow j| > \ell(L)\), where \(\ell(L)\) is the length of \(L\) (one less than the number of elements in a longest chain).
• Theorem 2.12. For every meet-irreducible \(m\), there exists a join-irreducible \(j \le m\) with \(|\uparrow j| = (|L|+1)/2\).

When \(|L|\) is even, \((|L|+1)/2\) is half-integral, so 2.12 is impossible for integer upper-set sizes. Thus any lattice meeting 2.12 must have odd order. A constructive search for lattices that pass both filters (a necessary step toward a Bouchard-compliant counterexample, or toward proving that none exist) therefore concentrates on odd-order lattices.

Note that Bouchard's conditions are necessary for a minimum-size counterexample specifically (his proofs consume minimality).

A natural first family is height-3 lattices: bottom, atoms, middle elements, and top. One might expect the cleanest specimens here — all join-irreducibles "low" (helping 2.7), a single middle rank for 2.12 targeting. We show that no height-3 lattice can co-satisfy, with the graded and JI-atomic structure emerging as a consequence of the conditions rather than an assumption. To the best of our knowledge, this is the first result establishing that Bouchard's Conditions 2.7 and 2.12 cannot be simultaneously satisfied at height 3.

The result grew out of a joint search with Isotopy (Claude-based) and Alethon (Grok-based) under stewardship of Sam White, in a shared working repository. Computational artifacts are available in frankl-verification. The lemma is due to Isotopy; Alethon independently checked the argument and the incidence constraints against an executable filter module implementing Bouchard's conditions.

2. Preliminaries

Let \(L\) be a finite lattice with bottom \(0\), top \(1\), and cardinality \(n = |L|\). Write \(\uparrow x = \{ y \in L : x \le y \}\) and \(\ell(L)\) for the length of \(L\) (the number of edges in a maximum-size chain).

An element \(j\) is join-irreducible if it covers exactly one element (equivalently: \(j = a \vee b\) implies \(j \in \{a,b\}\)). An element \(m\) is meet-irreducible if it is covered by exactly one element.

We say \(L\) is graded of height 3 if every maximal chain has exactly four elements \[ 0 \prec a \prec m \prec 1 \] (so \(\ell(L) = 3\)), and the elements partition into ranks \(0,1,2,3\).

We say \(L\) is atomic at the join-irreducibles (for short: JI-atomic) if every join-irreducible has rank 1 — i.e., the join-irreducibles are exactly the atoms. In a height-3 graded lattice, if every middle-rank element covers at least two atoms, then no middle element is join-irreducible, and the join-irreducibles are exactly the atoms.

Notation for the graded JI-atomic setting. After step (0) of the proof establishes these properties, we work in the following setting. \(L\) is a finite lattice with:

1. \(n = |L|\) odd and \(n \ge 5\);
2. \(L\) graded of height 3 (\(\ell(L) = 3\));
3. every join-irreducible is an atom (rank 1).

Write \(A\) for the set of atoms (so \(J = A\) under (3)), \(k = |A|\), and \(M\) for the set of rank-2 ("middle") elements, with \(m = |M|\). Then \[ n = 2 + k + m. \] For an atom \(a\), every element of \(\uparrow a\) other than \(a\) and \(1\) is a middle element containing \(a\), so \[ |\uparrow a| = 1 + r_a + 1 = r_a + 2, \] where \(r_a = |\{ \mu \in M : a < \mu \}|\) is the number of middle elements above \(a\).

3. Theorem 1 — Height-3 impossibility

Theorem 1 (Height-3 impossibility; Isotopy).
No finite lattice of height 3 satisfies both Bouchard's Theorem 2.7 and Theorem 2.12.

Proof

Let \(L\) be a finite lattice with \(\ell(L) = 3\), and assume for contradiction that both 2.7 and 2.12 hold.

(0) Structural consequences of 2.7 and 2.12 at height 3.
Three hypotheses one might impose — gradedness, atomic join-irreducibles, odd order — are in fact forced.

Odd order. 2.12 requires \(|\uparrow j| = (n+1)/2\) for some join-irreducible \(j\). Since \(|\uparrow j|\) is a positive integer, \(n\) must be odd.

All join-irreducibles are atoms. Any non-atom element \(x \ne 0, 1\) lies strictly above some atom, so the longest chain from \(x\) to \(1\) has length at most 1, giving \(|\uparrow x| \le 2\). If \(x\) were join-irreducible, 2.7 would require \(|\uparrow x| > 3\), a contradiction. If \(1\) itself were join-irreducible, \(|\uparrow 1| = 1 \le 3\), also contradicting 2.7. So \(J(L) = A\) (the atoms).

Effective gradedness. For any atom \(a\), 2.7 requires \(|\uparrow a| > 3\), so \(a\) cannot be a coatom (which would give \(|\uparrow a| = 2\)). Thus every atom lies below some element of \(L \setminus \{0, 1, \text{atoms}\}\) — the "middle" elements. Each middle element is covered only by \(1\) (by the depth argument above) and covers at least two atoms (otherwise it would be join-irreducible). Distinct middles are incomparable: if \(\mu_1 < \mu_2\) for middles \(\mu_1 \ne \mu_2\), then \(0 < a < \mu_1 < \mu_2 < 1\) (for any atom \(a < \mu_1\)) is a chain of length 4, contradicting \(\ell(L) = 3\). So the middles form an antichain. Let \(A\), \(k\), \(M\), \(m\), \(r_a\) be as in §2.

(i) Every middle element covers at least two atoms.
A middle element covering exactly one atom would be join-irreducible, contradicting \(J(L) = A\) established in (0).

(ii) Any two middle elements share at most one atom.
Suppose \(\mu_1, \mu_2 \in M\) (\(\mu_1 \ne \mu_2\)) both lie above distinct atoms \(a\) and \(b\). Then \(\mu_1 \wedge \mu_2 \ge a\) and \(\mu_1 \wedge \mu_2 \ge b\). Since distinct middles are incomparable (they form an antichain by the depth argument above), the meet is strictly below both, so it has rank 0 or 1 — that is, it is \(0\) or an atom. But no atom or \(0\) lies above two distinct atoms. Contradiction. Any two middles share at most one atom.
(Equivalently: the incidence structure \((A, M, \text{covers})\) is a partial linear space.)

(iii) Some atom realizes the 2.12 target.
By 2.12 and the existence of at least one meet-irreducible in a non-trivial lattice of this shape, there is a join-irreducible \(a\) — necessarily an atom — with \[ |\uparrow a| = \frac{n+1}{2}. \] Hence \(r_a = (n+1)/2 - 2 = (n-3)/2\).

(iv) Lower bound on \(k\).
The \(r_a\) middle elements containing \(a\) each contain at least one non-\(a\) atom (by (i)). No two of them share a non-\(a\) atom (by (ii), since they already share \(a\)). Therefore at least \(r_a\) distinct non-\(a\) atoms are required: \[ k - 1 \ge r_a = \frac{n-3}{2} \implies k \ge \frac{n-1}{2}. \]

(v) The target atom lies under every middle.
From \(n = 2+k+m\) and (iv), \[ m = n-2-k \le n-2-\frac{n-1}{2} = \frac{n-3}{2} = r_a. \] But \(r_a \le m\) always, so \(r_a = m\): atom \(a\) lies in all middle elements.

(vi)–(vii) Every other atom has replication number 1.
Since \(a\) lies in every middle, any two middles share \(a\), and by (ii) they share nothing else. Each non-\(a\) atom therefore appears in at most one middle. Every atom must appear below at least one middle: an atom covered directly by the top would create a maximal chain of length 2, violating gradedness. The \(m = k-1\) middle elements each need at least one non-\(a\) atom (by (i)), and the \(k-1\) non-\(a\) atoms each appear in at most one middle. So each non-\(a\) atom appears in exactly one middle, and for every atom \(x \ne a\), \[ r_x = 1, \qquad |\uparrow x| = 3. \]

(viii) Contradiction to 2.7.
Theorem 2.7 requires \(|\uparrow x| > \ell(L) = 3\) for every join-irreducible \(x\). But \(|\uparrow x| = 3\) for all \(x \ne a\). Contradiction. ∎

Remark (Star-in-disguise). The proof reveals that the 2.12-target atom \(a\) necessarily creates a star-like incidence structure: \(a\) is the hub, each middle element is a spoke, and every non-\(a\) atom is a leaf on exactly one spoke. Theorem 1 says that height-3 lattices satisfying 2.7 and 2.12 are, at the atom-to-middle incidence level, forced into a star configuration by the 2.12 requirement — and stars cannot co-satisfy with 2.7.

4. Remarks on scope

4.1 What the theorem closes

• All finite lattices of height 3, regardless of whether they are graded, regardless of where their join-irreducibles sit, and regardless of parity. The proof derives all of these structural properties from 2.7 and 2.12 alone.
• In particular, no height-3 lattice is a candidate for simultaneous passage of Bouchard 2.7 and 2.12, regardless of further filters (2.2–2.6, 2.9–2.11, 2.13).

4.2 What the theorem does not claim

• It does not assert that no finite lattice co-satisfies 2.7 and 2.12. That global statement is false: a height-4 witness exists (Section 5).
• It does not rule out height \(\ge 4\), non-graded lattices, or lattices with join-irreducibles above rank 1.
• It does not by itself prove Frankl's conjecture or that no Bouchard-compliant counterexample exists. It eliminates one constructive neighborhood.

4.3 Even order

If \(n\) is even, 2.12 fails for a simpler reason (non-integral target). The interesting case is odd \(n\), which the lemma addresses.

5. Theorem 2 — Height-4 sharpness witness

Theorem 2 (Existence at height 4; Fable exhaustive search, joint verification).
There exists a finite graded lattice \(L\) of height 4 co-satisfying both Bouchard's Theorems 2.7 and 2.12. The minimum odd cardinality among graded height-4 lattices with \(J(L) = \text{atoms}\) achieving this co-satisfaction is \(n = 13\).

The height-3 hypothesis in Theorem 1 is therefore essential for impossibility.

5.1 Enumeration

An exhaustive search over all graded lattices of height 4 whose join-irreducibles are exactly the atoms (covers only between adjacent ranks; every rank-2 element covers \(\ge 2\) atoms; every rank-3 element covers \(\ge 2\) rank-2 elements; the lattice property verified for all pairs) yields:

The \(n = 9\) class is empty for a structural reason requiring three bounds. First, \(k \ge 3\): with only 2 atoms, every rank-2 element covers both, so any two rank-2 elements share two atoms and their join fails. Second, \(s_2 \ge 3\): with \(s_2 = 2\) (and \(s_3 \ge 2\)), every rank-3 element covers both rank-2 elements, so the two rank-2 elements have two incomparable minimal upper bounds and their join fails — the same obstruction one rank up. Third, \(s_3 \ge 2\): with \(s_3 = 1\), the top covers exactly one element and is therefore join-irreducible, contradicting JI-atomicity. Together: \(n = 2 + k + s_2 + s_3 \ge 2 + 3 + 3 + 2 = 10\), so no \(n = 9\) lattice exists in this class.

5.2 The specimen

The unique (up to isomorphism) height-4 co-satisfying lattice at \(n = 13\):

Clean Hasse diagram (bot \(=0\), top \(=12\); \(\prec\) means covers):

rank 0:  0
rank 1:  1, 2, 3, 4            (all join-irreducible)
rank 2:  5, 6, 7, 8, 9
rank 3:  10, 11
rank 4:  12

0 ≺ 1,2,3,4
1 ≺ 5,6,7       2 ≺ 5,8,9       3 ≺ 6,8       4 ≺ 7,9
5 ≺ 10,11       6 ≺ 10           7 ≺ 11        8 ≺ 10        9 ≺ 11
10 ≺ 12         11 ≺ 12

Upper-set sizes: \(|\uparrow 1| = |\uparrow 2| = 7\), \(|\uparrow 3| = |\uparrow 4| = 5\). Target \((13+1)/2 = 7\).

Meet-irreducibles: \(M(L) = \{6, 7, 8, 9, 10, 11\}\). For 2.12, atom 1 (\(|\uparrow| = 7\)) witnesses meet-irreducibles \(6, 7, 10, 11\); atom 2 (\(|\uparrow| = 7\)) witnesses \(8, 9, 10, 11\). Every meet-irreducible is covered.

The escape from Theorem 1 is multi-rank middles, not multi-height join-irreducibles. Theorem 1's step (v) forces the 2.12-target atom into all middle elements when there is a single middle rank. Here, atom 1 achieves \(|\uparrow 1| = 7\) by covering 3 of 5 rank-2 elements (5, 6, 7). Through element 5 alone, atom 1 reaches both rank-3 elements (10 and 11) and hence the top. At height 3, covering 3 of 5 middles would leave 2 middles without a target-atom witness for 2.12. At height 4, the rank-3 layer provides a second opportunity: elements 10 and 11 are above atom 1 even though some rank-2 elements are not.

5.3 Bouchard scorecard

Bouchard's §2 contains fifteen numbered results (eight theorems, one lemma, six corollaries): 2.1–2.15. The specimen passes:

• 2.1–2.8, 2.10, 2.12–2.15 (13 conditions)

and fails:

• 2.9 (for every nonempty set \(M\) of meet-irreducibles, some join-irreducible \(j\) satisfies \(|\uparrow j \cap M| > |M|/2\) — fails on \(M = \{6, 9\}\): no join-irreducible lies below both, since the atom sets \(\{1,3\}\) and \(\{2,4\}\) are disjoint)
• 2.11 (the two-element instance of 2.9: every pair of meet-irreducibles has a common join-irreducible below both — fails on the same pair \(\{6, 9\}\))

Notably, the specimen passes 2.13 (\(\mathrm{Inc}(x)\) is never a chain for \(x \ne 0, 1\)). The 2.9/2.11 obstruction is a different geometry from 2.13: it concerns the overlap structure of atom sets below meet-irreducibles, not the comparability of incomparable elements.

The present paper does not claim a full Bouchard survivor — only joint 2.7\(\wedge\)2.12 at height 4, establishing sharpness of Theorem 1. Whether a lattice satisfying all 15 Bouchard conditions exists remains open.

6. Related search landscape (brief)

Outside the theorem's scope, product families such as \(M_3 \times C_3\) (order 15) pass 2.7 and fail only 2.12 among the Bouchard conditions in our module — a "closest miss" on a different axis (sole 2.12 failure vs. the \(n = 13\) specimen's 2.9/2.11 failure). Chain products \(C_a \times C_b\) unconditionally fail 2.12: for a join-irreducible in factor \(i\), \(|\uparrow j|\) is a multiple of \(|L|/|L_i|\), and \(\gcd(|L|/|L_i|, (|L|+1)/2) = 1\) for genuine products with \(\ge 2\) nontrivial factors. Partition lattices \(\Pi_4, \Pi_5\) pass 2.7 but miss 2.12 (and \(\Pi_4\) fails additional majority-coverage conditions). These data points shape search priorities; they are not needed for the theorems.

7. Conclusion

Sharp height boundary. Among finite lattices:

Theorem 1 is elementary lattice theory plus a partial-linear-space incidence constraint at the single middle rank; the contradiction is numerical and exact. The graded, odd-order, and JI-atomic hypotheses that one might impose are in fact forced by the combination of 2.7 and 2.12 at height 3, so the theorem carries no structural hypotheses at all beyond height. Theorem 2 shows co-satisfaction becomes possible at height 4, via multi-rank middles rather than elevated join-irreducibles, with \(n = 13\) as the provably minimal witness in the graded atomic class.

The remaining Bouchard failures at \(n = 13\) (conditions 2.9 and 2.11, concerning the overlap structure of atom sets below meet-irreducibles) identify the next obstruction for constructive search. Whether a lattice satisfying all 15 Bouchard conditions exists remains open and is the central question of the lattice approach to Frankl's conjecture.

Authorship and process

Acknowledgments. Rheon (ChatGPT Sol) independently verified Theorem 1, caught the n=9 error, observed that the odd-order and JI-atomic hypotheses are redundant, and provided a detailed review of the posted draft. Rheon was consulted by Sam White and acted as independent referee.

Steering level: collaborative (agent–agent math with human facilitation and independent verification). Research correspondence 2026-08-02–03; joint draft maintained in frankl-search/papers/height-boundary/.

Code / data: https://github.com/isotopyofloops/frankl-verification
Correspondence: math sprint thread between Isotopy and Alethon, 2026-08-02–03.

References

[1] C. Bouchard, On the lattice formulation of the union-closed sets conjecture, arXiv:2503.00277, 2025; Le Matematiche 81(1), 2026, pp. 153–165.

[2] Shared implementation notes and filter module: bouchard_filters.py, bouchard_joint_notes.md, https://github.com/isotopyofloops/frankl-verification (2026).

Appendix A. Filter implementations used for verification

In the shared module, 2.7 and 2.12 are checked as:

• 2.7: for all join-irreducibles \(j\), \(|\uparrow j| > \ell(L)\).
• 2.12: for all meet-irreducibles \(m\), some join-irreducible \(j \le m\) has \(2|\uparrow j| = n+1\) (integer comparison).

Upper sets are computed from the transitive closure of the covering relation. All cited scorecards use the corrected filter module. The \(n = 13\) specimen's lattice property was verified by checking all \(\binom{13}{2} = 78\) pairs for unique meets and joins.

Appendix B. Pair-lattice closed form (optional detail)

For the special case where middle elements are exactly all 2-subsets of a \(k\)-set (when that structure is a lattice), one has \(|\uparrow\mathrm{atom}| = k+1\) and \(n = k(k+1)/2 + 2\). Setting \(k+1 = (n+1)/2\) yields only trivial solutions \(k \in \{1,2\}\). The lemma covers this case and all other partial-linear-space middle layers without specializing the block size.